Question

Area of the curve
r = 2 − sin θ

Answer 1

i dont know it but here is a website it may help https://cims.nyu.edu/~kiryl/Calculus/Section_9.4--Areas_and_Lengths_in_Polar_Coordinates/Areas_and_Lengths_in_Polar_Coordinates.pdf

Answer 2

Are there bounds listed in the problem? Or is it just the area inside of this object? http://www.wolframalpha.com/input/?i=plot+r%3D2-sin%28theta%29

Answer 3

no boundaries were listed, although I was able to correctly choose the graph and it appears that x goes from -2 to 2 but I don't know how that could be since there is a trig function.

Answer 4

I believe the bound is from 0 to 2pi

Answer 5

Well I did it two different ways in polar coordinates and got two different answers.... trying to work it out now

Answer 6

Ok so I worked it out using the fact that the area element in polar coordinates is r dr d theta

Answer 7

So for a given figure:
\[ \int f(x,y)dx dy= \int f(r,\theta)r dr d\theta\]

Answer 8

I got 4pi as my answer

Answer 9

negative 4pi

Answer 10

Now blindy applying this trick and considering r going from 1 (smallest value it takes to 3 (|-3| the "largest" value it takes) and theta from 0 to 2 pi I got 16 pi.. which is the area of a circle of radius 4

Answer 11

but the radius i believe is 2 so then it would have to be 4pi

Answer 12

Im not sure how you got that... my other answer isn't 4 pi... Anyways I don't like that it works out like that because it CLEARLY isn't a circle the top segment is flat.... And no it isn't a circle of radius 2 since r is measured from the origin not from the center of the object

Answer 13

okay however when i tried 16pi it was incorrect

Answer 14

Yea know I know i wasnt finished hold on

Answer 15

sorry hold on I am just checking something

Answer 16

try 2 pi +8

Answer 17

that is incorrect

Answer 18

how many more shots? because my other answer is 8 pi +2

Answer 19

8pi+2 is also incorrect. I believe i have one more shot

Answer 20

Alright last shot I am taking into account the fact that area under the x-axis is negative with this one... so try -2pi-8

Answer 21

Incorrect no more tries.

Answer 22

Sorry man Im not sure what I am doing wrong everything seems to make sense... I broke the region up into 4 quarters and calculated the area of each (well only the top and bottom first quarter since the area above and below is twice each of these values since the object is symmetric with respect to the y-axis)

Answer 23

Does it list the correct answer?

Answer 24

Not until the actual assignment is due

Answer 25

Well that is kind of dumb how are you supposed to learn from a problem you got wrong unless you have something to work towards.

Answer 26

pellet I think that should be positive